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this is my first time on Stack Exchange so if I did something wrong please tell me.

I have a time series dataset.
There is an observation $(y,x)$ for each continuous time $t$.
Let’s say for each day since 2014-01-01, i.e. $t \in \left\{0, 1, 2, …, 108 \right\}$, where $108$ means today.

I ran a regression on the model
$y = \alpha + \beta x$
and found $\alpha$ and $\beta$ with $R^2 > 0.975$

My question is, can I approximate $\beta'$ in the model
$y' = \alpha' + \beta'x'$
where
$y'=\frac{y_{t+1}}{y_t}-1$
$x'=\frac{x_{t+1}}{x_t}-1$
without running regression on that model,

just by using $\hat{\beta'}= \frac{\beta \times x_{108}}{\hat{y_{108}}}$

where $\hat{y_{108}}$ is estimated using the original model.

I'm curious because I found that my $\hat{\beta'}$ is usually within 1% of $\beta'$

Even a simple Yes or No with help me a lot.

Thanks!

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1 Answer 1

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The answer is "it depends".

It depends on what is the true process which generated this data, aka DGP (data generation process). Your approximation will not hold in a general case.

I'll give you an example.

Let's assume the true DGP: $\ln y_t=\alpha+\beta \ln x_t+\varepsilon_t$, where $\varepsilon_t\sim\mathcal{N}(0,\sigma)$

Differenceing gives: $\Delta \ln y_{t+1}=\ln\frac{y_{t+1}}{y_{t}}=\beta\ln \frac{x_{t+1}}{x_t}+\Delta\varepsilon_{t+1}$

Since $\Delta \ln y_{t+1} \approx\frac{y_{t+1}}{y_{t}}-1=y_t'$, you get $y_t'=\beta x_t'+\Delta\varepsilon_{t}$

Now, if in your sample $x_t,y_t\sim 1$ then $\ln y_t\approx y_t-1$, , plugging this into DGP we get your model:

$y_t=\tilde{\alpha}+\beta x_t+\varepsilon_t$

So, my guess is that your $x_t,y_t$ happened to be close to 1, that's why it's working for you. Now that I wrote this all, I figured there's a simpler way to show this, but I'm too lazy to re-write it :)

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  • $\begingroup$ Thanks! What if $y = \alpha + \beta x$ is the true DGP, and $x_t \sim 100, y_t \sim 2,500,000 $? How will the answer change? $\endgroup$
    – base64
    Commented Apr 19, 2014 at 17:43
  • $\begingroup$ If I regress $\ln y=\alpha''+\beta'' \ln x$, I get $\beta'' = 0.90\beta$ $\endgroup$
    – base64
    Commented Apr 19, 2014 at 17:53
  • $\begingroup$ What are the variances? $\endgroup$
    – Aksakal
    Commented Apr 19, 2014 at 19:26
  • $\begingroup$ $\sqrt{Var(x)} = 20, \sqrt{Var(y)} = 500,000$ $\endgroup$
    – base64
    Commented Apr 19, 2014 at 19:48
  • $\begingroup$ You got lucky. It turns out that in your case $\Delta x_t/x_t\sim\Delta y_t/y_t$, look at the ratios of your dispersions and the variable levels. $\endgroup$
    – Aksakal
    Commented Apr 20, 2014 at 13:57

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